题意:给一个无向图,每条边上都有容量的限制,要求求出给定起点和终点的最大流. 思路:每条无向边就得拆成2条,每条还得有反向边,所以共4条.源点汇点已经给出,所以不用建了.直接在图上跑最大流就可以了. #include <bits/stdc++.h> #define LL long long #define pii pair<int,int> #define INF 0x7f7f7f7f using namespace std; ; ; int s, t; int path[N],…
1153 - Internet Bandwidth PDF (English) Statistics Forum Time Limit: 2 second(s) Memory Limit: 32 MB On the Internet, machines (nodes) are richly interconnected, and many paths may exist between a given pair of nodes. The total message-carrying cap…
Time limit: 3.000 seconds限时3.000秒 Problem问题 Given a graph (V,E) where V is a set of nodes and E is a set of arcs in VxV, and an ordering on the elements in V, then the bandwidth of a node v is defined as the maximum distance in the ordering between v…
https://vjudge.net/problem/UVA-820 题意: 给出所有计算机之间的路径和路径容量后求出两个给定结点之间的流通总容量. 思路: 裸的最大流问题.注意有个比较坑的地方,最后需要多输出一个空行,否则会wa. #include<iostream> #include<algorithm> #include<cstring> #include<queue> using namespace std; ; int n, m; int s, e…